We have three equations, one for the front entrance and two for the support beams.
Front Entrance & Beams
y=- 110(x-10)^2+20 & ly=- x+10 y=x-10
We want to find the points where the parabola and the lines intersect. To find these points, we need to solve two systems of equations.
SE_1: y=- 110(x-10)^2+20 & (I) y=- x+10 & (II)
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SE_2: y=- 110(x-10)^2+20 & (I) y=x-10 & (II)
Using the Zero Product Property (ZPP), we can get two solutions for x.
x(x-30)=0 ZPP ⟶ lx=0 x=30
Next, we will substitute the x-values into the linear equation and find the y-values.
x
- x+10
y=- x+10
0
- ( 0)+10
10
30
- 30+10
- 20
Since x and y represent the lengths and substituting 30 for x resulted in a negative y-value, we ignore the solution ( 30, - 20). Therefore, the beam with equation y=- x+10 and the parabola intersect at the point (0,10).
Solving the Second System
We will follow the same steps to solve the second system. Let's start by substituting the equivalent expression for y in Equation (II) into Equation (I).
We can use the ZPP again to find the two solutions for x.
(x-20)(x+10)=0 ZPP ⟶ lx=20 x=- 10
Since x represents the length, we ignore x=- 10. Let's substitute the other x-value into the linear equation and find the y-value.
x
x-10
y=x-10
20
20-10
10
The beam with equation y=x-10 and the parabola intersect at the point (20,10). As a result, the support beams and the parabola intersect at the points (0,10) and (20,10).
Alternative Solution
Solve by Graphing
We can solve the systems of equations by graphing. SE_1: y=- 110(x-10)^2+20 & (I) y=- x+10 & (II)
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SE_2: y=- 110(x-10)^2+20 & (I) y=x-10 & (II)
Let's first draw the parabola. Since the equation is in standard form, we can identify its vertex. y= - 1/10(x- 10)^2+ 20
We see that the vertex of the parabola is ( 10, 20) and the axis of symmetry is x=10. Its y-intercept is (0,10). y& =- 1/10(0-10)^2+20=10
Now, let's plot the vertex and y-intercept. Then, we will reflect the intercept across the axis of symmetry.
We will sketch the parabola which passes through the three points
Finally, we will draw the beam equations. The equation y=x-10 is a translation of y=x down by 10 units, and the equation y=- x + 10 is a translation of y=- x up by 10 units.
We see that the support beams meet the parabola at the points (0,10) and (20,10).
